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Dmitry
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Post subject: modality of singularities Posted: Wed Jun 18, 2008 2:20 pm |
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I haven't found a command to calculate the modality of a singular point. In some simple cases this can be done by running classify(f);
Guess this works for some lowest cases only. But it does not recognize (??) e.g. the case (x^5+y^6)
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Guest
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Post subject: Posted: Fri Jun 20, 2008 5:59 pm |
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Computing the modality is a non-trivial task and there is no general algorithm available besides the classification, except for plane curve singularities. Moreover, we have to distinguish the modality for right- and contact-equivalence (the latter is even more complicated than the first).
For curve singularities we can compute the right-modality with the help of an embedded resolution, cf. G.-M. Greuel, C. Lossen, E. Shustin: Introduction to Singularities and Deformations. Springer Verlag, Berlin, Heidelberg, New York (2006), p. 373, Remarks and Exercises. Or by the algorithm developed in A. Campillo, G.-M. Greuel, C. Lossen: Equisingular Calculations for Plane Curve Singularities. J. Symb. Comput. 42 (2007), 89-114. The latter is implemented in Singular in equising.lib.
If mod denotes the right modality of an isolated hypersurface singularity f then mod = dimension of the mu-constant stratum of f in the base of the semiuniversal deformation of f, where mu is the Milnor number of f. Hence, if f is semi-quasihomogeneous or Newton-nondegenerate, it can be computed with the help of the Newton diagram of f.
A discussion of this with examples about how to use the SINGULAR library equising.lib to compute the modality or, equivalently, the codimension tau_es of the mu-constant stratum for arbitrary plane curve singularities can be found at the above mentioned place too.
Example: LIB "equising.lib"; ring r = 0,(x,y),ds; poly f = x5+y6; milnor(f) - tau_es(f);
// -> 3 i.e. the right-modality is 3 (printlevel = 1; shows additional information) [/quote]
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Guest
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Post subject: Posted: Sat Jun 21, 2008 6:58 am |
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Dmitry
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Post subject: Something is strange with \tau^{es} !! Posted: Sat Jun 21, 2008 8:23 am |
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The command poly f = (x2+y2)*(y2-(x-y)^3)*(x^3+y9); tau_es(f);
returns: tau_es(f)=37
while the command poly f = ((x+y)^2+y2)*(y2-x3)*((x+y)^3+y9); tau_es(f);
returns: tau_es(f)=23
note, however, that the polynomials differ by just a linear change of variables x->x+y
A bug in Singular?
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greuel
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Post subject: Posted: Mon Jun 23, 2008 2:00 am |
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Joined: Mon Aug 29, 2005 9:22 am Posts: 41 Location: Kaiserslautern, Germany
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Indeed, this is a bug. It will (hopefully) be fixed in the next version of SINGULAR. So far tau_es works only correct in the Newton nondegenerate case.
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hannes
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Post subject: bug fix Posted: Tue Jul 08, 2008 4:00 pm |
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Joined: Wed May 25, 2005 4:16 pm Posts: 275
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Guest
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Post subject: thanks! Posted: Thu Jul 10, 2008 7:25 am |
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